# Math Help - Limits question

1. ## Limits question

Limit question on a midterm that I have and still haven't the slightest idea on how to complete

Find all Values of a and b such that $\lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1$

all I could figure out was that if you use L'Hopitale's rule you would get

$\lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1$

but I have no idea how to use this to solve for any variables

thank you

2. Hello,

Why not using a second time l'Hôpital's rule ?

3. because it would be (0/0) ?

4. Hm ok, I just don't know correctly the conditions for applying this rule

5. Hi

Another approach : $\lim_{x\to 0} \frac{\sin(2x)-\sin(0)}{2x-0}=\cos 0 = 1$
Hence, $\lim_{x \to 0} \frac{3}{x}\frac{\sin(2x)}{x}+a-\frac{b}{x}=\lim_{x \to 0} \frac{6}{x} \frac{\sin(2x)}{2x}+a-\frac{b}{x}=\lim_{x\to 0}\frac{6-b}{x}+a=1$
When is this possible ?

6. Originally Posted by akhayoon
Limit question on a midterm that I have and still haven't the slightest idea on how to complete

Find all Values of a and b such that $\lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1$

all I could figure out was that if you use L'Hopitale's rule you would get

$\lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1$

but I have no idea how to use this to solve for any variables

thank you
L'hopital it again if you wish...it works...but look...you have two equations with two variables...solve for them using simulataneous equations

7. Originally Posted by akhayoon
Limit question on a midterm that I have and still haven't the slightest idea on how to complete

Find all Values of a and b such that $\lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1$

all I could figure out was that if you use L'Hopitale's rule you would get

$\lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1$

but I have no idea how to use this to solve for any variables

thank you
Obviously you require another indeterminant form. This will happen if b = -6. So then you have:

$\lim_{x \rightarrow 0} \frac{6 \cos(2x) + 2ax - 6}{2x}=1$.

Apply l'Hopital again:

$\lim_{x \rightarrow 0}\frac{-12 \sin (2x) + 2a}{2}=1$.

In other words, $\frac{2a}{2} = 1 \Rightarrow a = 1$.

8. Originally Posted by akhayoon
Limit question on a midterm that I have and still haven't the slightest idea on how to complete

Find all Values of a and b such that $\lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1$

all I could figure out was that if you use L'Hopitale's rule you would get

$\lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1$

but I have no idea how to use this to solve for any variables

thank you
All your troubles would have been avoided of course if you hadn't run to the l'Hospital with a sore thumb. Try substituting the series expansion around x = 0 for sin(2x):

$\Rightarrow \lim_{x \rightarrow 0} \frac{3 (2x - [2x]^3/6 + ...) + ax^{2} - bx}{x^2}=1$

$\Rightarrow \lim_{x \rightarrow 0} \frac{(6x - 4x^3/3 + ...) + ax^{2} - bx}{x^2}=1$

$\Rightarrow \lim_{x \rightarrow 0} \frac{(6 - b) x + ax^{2} - 4x^3/3 + ...}{x^2}=1$

$\Rightarrow \lim_{x \rightarrow 0} \frac{(6 - b) + ax - 4x^2/3 + ...}{x} = 1$

There's a problem unless 6 - b = 0 => b = 6. Then:

$\Rightarrow \lim_{x \rightarrow 0} \frac{ax - 4x^2/3 + ...}{x} = 1$

$\Rightarrow \lim_{x \rightarrow 0} \left(a - \frac{4x}{3} + ...\right) = 1$

Therefore a = 1.